Solve the following linear first order Pde with the characteristic curves method

Question

Solve the following PDE $$ y \cdot \frac{\partial}{\partial x}u+ \frac{\partial}{\partial y}u=x \\ u(x,0)=sin(x).$$ I cannot find a solution. Firstly I have calculated the flux $\Phi$ of the vector field which is $$ \Phi(t,(x,0))=\begin{cases} x(t)=\frac{(t+c)^2}{2}+d \\ y(t)=t+c \end{cases},$$ with $c=0$ because I want my integral curves to intersect the $x$ axis in $t=0.$ If a solution $u$ exists the function $\hat{u}=u \circ \Phi$ has to solve the following Cauchy problem $$ \begin{cases} \frac{\partial}{\partial t}\hat{u}(t,(x,0))=x(t)= \frac{(t)^2}{2}+d \\ \hat{u}(0,(x,0))=sin(x) \end{cases} $$ So we must have $\hat{u}(t)=\frac{t^3}{6}+dt+a$ where $a=sin(x).$ At a certain time $t_1$ the integral curve starting at $(x,0)$ has to encounter the point $(x,y)$ obtaining $$ \begin{cases} x =\frac{t_1^2}{2}+d \\ y=t_{1}. \end{cases}$$ Finally, $$u(x,y)=\hat{u}(t_1) \\ = \frac{y^3}{6}+(x-\frac{y^2}{2})y+sin(x),$$ but this is not the solution. Where did I go wrong?

Answer 1

$$ y \cdot \frac{\partial}{\partial x}u+ \frac{\partial}{\partial y}u=x $$ https://en.wikipedia.org/wiki/Method_of_characteristics $$\frac{dx}{y}=\frac{dy}{1}=\frac{du}{x}$$ A first characteristic equation from $\frac{dx}{y}=\frac{dy}{1}$ : $$x-\frac12y^2=c_1$$ A second characteristic equation from $\frac{dy}{1}=\frac{du}{x}=\frac{du}{c_1+\frac12y^2}\quad\implies\quad du=(c_1+\frac12y^2)dy$ $$u-c_1y-\frac16y^3=c_2$$ General solution of the PDE : $c_2=F(c_1)=u-c_1y-\frac16y^3=F(x-\frac12y^2)$

$F$ is an arbitrary function. $$u(x,y)=(x-\frac12y^2)y+\frac16y^3+F(x-\frac12y^2)$$ $$\boxed{u(x,y)=xy-\frac13y^3+F(x-\frac12y^2)}$$ Condition : $u(x,0)=\sin(x)=0.x-0+F(x-0)$ $$F(x)=\sin(x)$$ The function $F$ is determined. We put it into the above general solution where the argument is $(x-\frac12y^2)$. $$\boxed{u(x,y)=xy-\frac13y^3+\sin(x-\frac12y^2)}$$